Aksum University Turbo Machines Lab Manual

1
AKSUM UNIVERSITY
COLLEGE OF ENGINEERING AND TECHNOLOGY
DEPARTMENT OF MECHANICAL ENGINEERING
TURBO MACHINES & MACHINES LAB MANUAL
Compiled by
A.Syed Bava Bakrudeen,
Associate Professor,
Mechanical Department
Aksum University, Ethiopia
LIST OF EXPERIMENTS
1. Pelton turbine
2. Reaction turbine (Efficiency Test)
3. Axial turbine
4. Reaction turbine (Velocity Test)
5. Centrifugal pump
6. Piston pump
7. Gear pump
8. Axial fan
9. Radial fan
10. Tubular heat exchanger (Parallel flow)
11. Tubular heat exchanger (Counter flow)
12. Air compressor
13. Multipurpose air duct.

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1. PELTON TURBINE
THEORY: Pelton wheel turbine is an impulse turbine, which is used to act on
high loads and for generating electricity. All the available heads are classified in to
velocity energy by means of spear and nozzle arrangement. Position of the jet
strikes the knife-edge of the buckets with least relative resistances and shocks.
While passing along the buckets the velocity of the water is reduced and hence an
impulse force is supplied to the cups which in turn are moved and hence shaft is
rotated.
AIM:
To determine the efficiency of the given pelton turbine using constant
volume method.
APPARATUS REQUIRED:
1. Pelton turbine setup.
2. Collecting tank
3. Stop watch
4. Meter scale
FORMULA:
1. Mechanical Power = Pmech = Mω = Mx 2xπxn/60
M=Moment in N/m.
n = Speed in rpm.
2. Hydraulic Efficiency = Phyd = pxV
p = Pressure in N/m2
.
V= Volume flow rate in m3
/Sec (Discharge)
(1liter =1000 cm3
= 0.001 m3
) (1bar = 1x105
N/m2
)
(1liter/min =16.67x10-6
m3
/Sec) (1liter = 1kg)
3. Efficiency η = Pmech x100/ Phyd
PROCEDURE:
1. Tare values in the system diagram
2. Select "Measurement Diagram" in the program.

3
3. Enable new series of measurements. Make settings for the measurements file.
4. Open the brake fully with the adjusting screw(8).
5. Open valve (4) fully to create the maximum volume flow.
6. Change the pump to "Pressure Control" and specify a control pressure. This
corresponds to the constant height difference of a real turbine to the storage
lake. With the pumps used it is possible to achieve pressures up to about 3.8 bar
with the needle nozzle fully opened. Higher pressures are only possible by
throttling the volume flow. We recommend using this pressure range. The
measured values obtained by GUNT reach 4 bar.
7. Wait until the operating point is reached. In this case the observation of the
water jet from the blade is particularly relevant for the subsequent explanation.
Then record measurements (the current measurement data set is written to the
measurements file). The program is now ready for the next measurement.
8. a) Use the adjusting screw on the brake (8) to increase the brake torque. The
turbine's rotational speed and torque change. The change depends on the desired
number of measurement points. Meaningful characteristics are often obtained
with 5 to 6 measurement points.
b) The volume flow can still be varied at each braking position. To do this, the
nozzle is closed gradually from the initial fully open position.
9. Repeat steps 7 and 8 as often as needed until the turbine's rotational speed has
fallen to zero and tabulate the value.
GRAPHS:
i Torque (M) Vs Speed (n) (Ordinary Graph)
RESULT:
1. Maximum efficiency of the turbine = ……………………… %
2. Speed corresponding to maximum efficiency = ………….. rpm
3. Power corresponding to maximum efficiency = …………..W.
4. Discharge corresponding to maximum efficiency = ………….. L/min.

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TABULATION:
S.
No
Torque
(M)
Ncm
Pressure
(p) bar
Speed(n)
1/min
Volume
Flow V
(L/min)
Mechanical
Power
(Pmech)
W
Hydraulic
Power
(Phyd)
W
Efficiency
(η) %
1. 22.8 3.0 2260 22 53.96 110 49.05
MODEL CALCULATIONS:
1. Phyd = pxV = 3x105
x22x0.001/60 = 110 W
2. Pmech = Mω = Mx 2xπxn/60 = 22.8x10-2
x2xπx2260/60 = 53.96 W.
3. η = Pmech x100/ Phyd= 53.96/110x100 =49.05
Fig1.Pelton Turbine

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2. REACTION TURBINE (EFFICIENCY TEST)
THEORY: Reaction turbines are acted on by water, which changes pressure as it
moves through the turbine and gives up its energy. They must be encased to
contain the water pressure (or suction), or they must be fully submerged in the
water flow. Newton's third law describes the transfer of energy for reaction
turbines.Most water turbines in use are reaction turbines and are used in low
(<30 m or 100 ft) and medium (30–300 m or 100–1,000 ft) head applications. In
reaction turbine pressure drop occurs in both fixed and moving blades. It is largely
used in dam and large power plants
AIM:
To determine the efficiency of the given reaction turbine under constant
pressure condition.
APPARATUS REQUIRED:
1. Francis turbine setup.
2. Collecting tank
3. Stop watch
4. Meter scale
FORMULA:
1. Mechanical Power = Pmech = Mω = Mx 2xπxn
M=Moment in N/m.
n = Speed in rpm.
(1liter =1000 cm3
= 0.001 m3
) and (1bar = 1x105
N/m2
)
.
/Sec (Discharge)
3. Efficiency = Pmech x100/ Phyd
PROCEDURE:

6
4. Open the brake fully with the adjusting screw(8).
5. Open valve (4) fully to create the maximum volume flow.
6. Change the pump to "Pressure Control" and specify a control pressure. Pressure
up to about 3 bar can be achieved with the pump used. Higher pressure cannot
be maintained over the entire operating range due to increasing the volume
flow.
7. Wait until the operating point is reached. Then record measurements (the
current measurement data set is written to the measurements file). The program
is now ready for the next measurement.
8. The brake torque is increased via the adjusting screw (7). The torque is varied
depend up on the desired number of measurement points. Meaningful
characteristics are often obtained with 10 measurement points.
9. Repeat steps 7 and 8 as often as needed until the turbine's rotational speed has
fallen to zero and tabulate the value.
GRAPHS:
RESULT:

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TABULATION:
S.
No
Torque
(M)
Ncm
Pressure
(p) bar
Speed
(n)
1/min
Volume
Flow
(V)
(L/min)
Mechanical
Power
(Pmech)
W
Hydraulic
Power
(Phyd)
W
Efficiency
(η) %
1. 7.9 3 8848 28.3 73.2 141.5 51.73
MODEL CALCULATIONS:
1. Phyd = pxV = 3x105
x28.3x0.001/60 = 141.5 W
2. Pmech = Mω = Mx 2xπxn/60 = 7.9x10-2
x2xπx8848/60 = 73.2 W.
3. η = Pmechx100/Phyd = (73.2/141.5)x100 = 51.73%.
Fig.2 Reaction Turbine

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3. AXIAL TURBINE
THEORY: If the water flows parallel to the axis of the rotation of the shaft, the
turbine is known as axial flow turbine. If the head at the inlet of the turbine is the
sum of pressure energy and kinetic energy and during the flow of water through
runner a part of pressure energy is converted into kinetic energy, the turbine is
known as reaction turbine.
AIM:
To determine the efficiency of the given axial flow axial turbine under
constant speed and pressure.
APPARATUS REQUIRED:
1. Kaplan turbine setup.
2. Collecting tank
3. Stop watch
4. Meter scale
FORMULA:
1. Mechanical Power = Pmech = Mω = Mx 2xπxn
M=Moment in N/m.
n = Speed in rpm.
(1 Millibar = 0.001 bar)
.
/Sec (Discharge)
3. Efficiency = Pmech x100/ Phyd
PROCEDURE:
4. Open the brake fully with the adjusting screw.
5. Open control valve V1 to obtain maximum upstream pressure
6. Switch on the pump.

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7. Wait until the operating point is established. Then record the measuring values
(the current measurement data set is written to the measurement file). The
program is now ready for next measurement.
8. Increase the turbine torque using the adjusting screw. The torque variation is
depending on the number of measuring points chosen. Meaningful
characteristics are often obtained with 5 to 6 measuring points.
10.Repeat steps 7 and 8 as often as needed until the adjusting screw is at the stop.
11. Save the measurement file.
12.Alternatively, further series of measurement can be recorded at reduced
upstream pressure. To do so, reduce the pressure at control valve V1 and repeat
the measurement from step 6 onwards.
GRAPHS:
RESULT:

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TABULATION:
S.
N
o
Torque (M)
(Ncm)
Pressure
(p)
(mbar)
Speed
(n)
(1/min)
Volume
Flow (V)
( L/min)
Mechanical
Power
(Pmech) (W)
Hydraulic
Power
(Phyd) (W)
Efficiency
(η) (%)
1. 34.8 890 3490 125 127.2 185.42 68.6
MODEL CALCULATIONS:
1. Phyd = pxV = 0.89x105
x125x0.001/60 = 185.42 W
2. Pmech = Mω = Mx 2xπxn/60 = 34.8x10-2
x2xπx3490/60 = 127.2 W.
3. η = Pmechx100/ Phyd = (127.2 /185.42)x100 =68.6%
Guided Wheel Impeller
Fig3. Axial Turbine

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4. REACTION TURBINE ( VELOCITY TEST)
THEORY:
Absolute velocity (c): The velocity that the water jet has at the outlet from the
turbine compared to the environment.
Circumferential velocity (u): The velocity of the impeller. Since this depends on
the flow of the exiting water jet, it is the velocity at the diameter of the outlet
nozzle.
Relative velocity (w): It is the velocity corresponds to the velocity of the flow
relative to the nozzle. It is calculated by adding circumferential velocity (u) and
absolute velocity (c).
AIM:
To find the relative velocity and circumferential velocity of the francis
turbine and determine the correlation between relative velocity and volume flow
rate of the given turbine.
APPARATUS REQUIRED:
1. Francis turbine setup.
2. Collecting tank
3. Stop watch
4. Meter scale
FORMULA:
1. Circumferential Velocity = u2 = πdn/60
d= Diameter of nozzle = 52 mm.
n = Speed in rpm.
2. Relative velocity = W2 = √(((Pux2)/ρ)+ u2
2
)
pu = Net Pressure in N/m2
.
ρ = Density of water = 1000 kg/m3
.
PROCEDURE:
4. Loosen off adjusting screw to open the brake.

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5. Let the pump run to100% power (This ensures that air in the flow section does
not affect the measured values).
6. Wait until the operating point is reached. Then record measurements (the
current measurement data set is written to the measurements file). The program
is now ready for the next measurement.
7. The capacity of the pump is retracted one step. The power is varied depending
on the desired number of measurement points. Meaningful characteristics are
often obtained with 5 to 6 measurement points
8. Repeat steps 6 and 7 as many times as needed until there is no more rotational
speed at the turbine.
9. Save the measurements file.
GRAPHS:
i Relative Velocity Vs Volume flow.
OBSERVATION: Diameter of nozzle (d) = 52 mm.
RESULT:
1. The maximum relative velocity = ________________ m/s.
2. Net pressure at inlet = _____N/m2
at maximum relative velocity
3. The relative velocity is directly proportional to volume flow rate.

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TABULATION:
S.
No
Pressu
re (p)
bar
Speed
(n)
1/min
Volume
Flow (V)
L/min
Circumferential Velocity
(u2) (m/s)
Relative velocity
(W2)
(m/s)
1 3.18 19894 46.7 54.16 59.8
MODEL CALCULATIONS:
1. u2 = πdn = πx0.052x19894/60 = 54.16 m/s
2. W2 = √(((Pux2)/ρ)+ u2
2
) = √(((3.18x105
x2)/1000)+ 54.22
)= 59.8 m/s
W
C
u2
Fig 4. Velocities in reaction turbine
C- Absolute velocity at outlet (m/s)
W= Relative velocity (m/s)
u2= Circumferential velocity (m/s)

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5. CENTRIFUGAL PUMP
THEORY: Centrifugal pumps are a sub-class of dynamic axisymmetric work-
absorbing turbomachinery. Centrifugal pumps are used to transport fluids by the
conversion of rotational kinetic energy to the hydrodynamic energy of the fluid
flow. The rotational energy typically comes from an engine or electric motor. The
fluid enters the pump impeller along or near to the rotating axis and is accelerated
by the impeller, flowing radially outward into a diffuser or volute chamber
(casing), from where it exits.
AIM:
To determine the characteristics of a centrifugal pump under constant speed
and varying the discharge method.
APPARATUS REQUIRED:
1. Centrifugal pump setup.
2. Collecting tank
3. Stop watch
4. Meter scale
FORMULA:
1. Hydrualic Power Phyd= ∆P xV in Watts.
∆P = P2-P1=Difference in pressure in N/m2
.
V = Volume flow rate (Discharge or Actual discharge) in m3
/sec.
(1L/min = 0.001 m3
/min) (1 min = 60 sec) (1 L/min = 16.66x10-6
m3
/sec)
(1bar = 1x105
N/mm2
)
2. Head H = ∆P/ ρxg in meter
ρ= Density of water=1000 kg/m3
.
g = Specific gravity =9.81 m/s2
.
(1L/min = 0.001 m3
/min)
3. Efficiency = Phyd/Pel.
Phyd = Hydrualic Power in W
Pel = Electrical Power in W
4. Specific speed Ns = n√V/ (H)3/4
(unitless)

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PROCEDURE:
1. Bleed the pump demonstrator.
2. Open valve V2 fully.
3. Use the Tare button to calibrate to zero.
4. Leave pump to turn to n=_________ rev/min.
5. Record measuring values of the suction pressure (P1), the pump outlet
pressure (P2), hydraulic and electrical power and volume flow (V).
6. Reduce the volume flow bit by bit by gradually closing valve V2 and take
the measurements according to point 5.
7. Repeat steps 5 and 6 until the volume flow is completely throttled.
8. Additional curves can be recorded with different rotational speed.
GRAPHS:
i Volume flow rate (V2) Vs Hydraulic Power (Phyd)
ii. Volume flow rate (V2) Vs Electrical Power (Pel)
iii. Volume flow rate (V2) Vs Efficiency (η)
RESULT:
1. Maximum efficiency of the pump = ……………………… %
2. Discharge corresponding to maximum efficiency = ………….. m3
/Sec.
3. Input power corresponding to maximum efficiency = …………..W.
4. Head corresponding to maximum efficiency = ………….. meter.

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TABULATION:
S
.
N
o
Rotation
al speed
(n) (Rev
/ min)
Volume
flow
(V)
(L/min)
Suction
Pressure
(P1) (bar)
Discharg
e
Pressure
(P2) (bar)
Temp
eratu
re (T)
(0
C)
Electric
al
Power(
Pel)
(W)
Pressure
Differen
ce (∆P)
(bar)
Head
(H)
(M)
Hydrau
lic
Power
(Phyd)
(W)
Efficien
cy (η)
(%)
Specifi
c speed
(Ns)
-
1 3300 25.2 -0.0056 0.865 25 343 0.8706 8.875 36.56 10.66 13.15
MODEL CALCULATIONS:
1. Phyd= ∆P xV = 0.8706X105
x 25.2x0.001/60 = 36.56
2. Head H = ∆P/ ρxg = 0.8706X105
/(1000X9.81) = 8.875
3. Efficiency = Phydx100/Pel = 36.56x100/343=10.66%.
= 3300 x √(25.2x0.001/60 )/ (8.875)3/4
= 13.15
Fig6. Schematic diagram of centrifugal pump
EI1 Energy input Pel of the pump B1 Water tank
FI1 Volume flow P1 Centrifugal pump
PI1 Pressure p1 upstream of the pump V1 Valve to throttle the suction side
PI2 Pressure p2 downstream of the pump V2 Valve to throttle the pressure side
TI1 Water temperature V3 Outlet valve

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6. PISTON PUMP
THEORY: Reciprocating is a positive displacement pump in which the liquid is
sucked and then it is actually pushed or displaced due to the thrust exerted on it by
a moving member, which results in lifting the liquid to the required height These
pumps usually have one or more chambers which are alternatively filled with the
liquid to be pumped and then emptied again As such the discharge of liquid
pumped by these pumps almost wholly depends on the speed of the pump. It is
widely used in Automobile Service Stations and Chemical Industries.
AIM:
To conduct the performance test on piston pump
APPARATUS REQUIRED:
1. Piston pump setup.
2. Collecting tank
3. Stop watch
4. Meter scale
FORMULA:
1. Hydrualic Power Phyd= ∆P xV in Watts.
.
/sec.
(1L/min = 0.001 m3
/min) (1 min = 60 sec) (1 L/min = 16.66x10-6
m3
/sec)
(1bar = 1x105
N/mm2
.)
2. Indexed Work Pind = Wind x60 / n
Pind = Indexed work in joules.
.
.
5. Qt = Theoretical discharge = 2xLxAxn/60 in m3
/sec

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Where Area of the piston A= (π/4) x d2
in m2
.
d= diameter of the piston =32mm
L= Stroke length of the piston = 15mm
n = Motor speed in rpm.
6. Percentage of slip = (Qt – V)x100/Qt.
PROCEDURE:
1. Open the throttle valve and suction-side throttle valve fully.
2. Check whether the air cushion in the air vessel fills approx. half the useable
volume.
3. Tare values in the system diagram.
4. Set overflow valve to maximum set point.
5. Switch on the piston pump, set speed to 100min-1
.
6. Gradually close the throttle valve (7) until the supply pressure p2 is approx.
2 bars.
7. Save a screenshot of the system diagram in a file.
GRAPHS:
OBSERVATIONS: d= diameter of the piston =32mm
L= Stroke length of the piston = 15mm
RESULT:
/Sec.
5. Percentage of slip corresponding to maximum efficiency=………..

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TABULATION:
S
.
N
o
Rotat
ional
speed
(n)
(Rev
/ min)
Volume
flow
(V)
(L/min)
Suctio
n
Pressu
re (P1)
(N/m2
)
Disch
arge
Pressu
re (P2)
(N/m2
)
Elect
rical
Powe
r (Pel)
(W)
Index
ed
powe
r (Pin)
(W)
Index
work
in (J)
Wind
Pressur
e
Differe
nce
(∆P)
(bar)
Hea
d
(H)
(M)
Hydrau
lic
Power
(Phyd)(
W)
Effic
ienc
y (η)
(%)
Theoretica
l
Discharge
(Qt) (m3
/s)
% of
Slip
1 94 1.03 -0.08 1.93 55.2 3.6 2.23 2.01 20.5 3.45 6.25 3.8 x10-5
53.4
MODEL CALCULATIONS:
1. Phyd= ∆P xV = 2.01x105
x1.03x0.001/60 =3.45 W
2. Wind = Pind x60 / n = 3.6x60/94=2.23 J
3. H = ∆P/ (ρxg) =2.01x105
/ (1000x9.81) = 20.5 m
4. η=Phyd x100/Pel = 3.45x100/55.2 = 6.25%
5. Qt = 2xLxAxn/60 = 2x 0.015 x(π/4)x 0.0322
x 94/60 = 3.8 x10-5
m3
/Sec.
6. % slip = (Qt – V)x100/Qt.
= (3.8x10-5
– (1.03x16.67x10-6
))x100/ 3.8x10-5
= 53.46%
Fig 6. Piston pump

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7. GEAR PUMP
THEORY: A rotary gear pump consists essentially of two intermeshing spur gears
which are identical and which are surrounded by a closely fitting casing. One of
the pinions is driven directly by the prime mover while the other is allowed to
rotate freely. The fluid enters the spaces between the teeth and the casing and
moves with the teeth along the outer periphery until it reaches the outlet where it is
expelled from the pump. Each tooth of the gear acts like a piston or plunger of on
reciprocating pump and hence the pump can be termed a positive displacement
pump. Gear pump is widely used for cooling water and pressure oil to be supplied
for lubrication to motors, turbine, machine tools etc.
AIM:
To determine the characteristics of a gear pump under constant speed and
varying the discharge and obtain the best-driven conditions by drawing the
performance curves.
APPARATUS REQUIRED:
1. Gear pump setup.
2. Collecting tank
3. Stop watch
4. Meter scale
FORMULA:
1. Hydrualic Power Phyd= HxV in Watts.
.
/sec.
(1L/min = 0.001 m3
/min) (1 min = 60 sec) (1 L/min = 16.66x10-6
m3
/sec)
(1bar = 1x105
N/mm2
)
.
.

21
(unit less)
PROCEDURE:
1. Bleed the pump demonstrator.
2. Open valve V2 fully.
3. Use the Tare button to calibrate to zero.
4. Leave pump to turn to n=_________ rev/min.
5. Record measuring values of the suction pressure (P1), the pump outlet
pressure (P2), hydraulic and electrical power and volume flow (V).
6. Reduce the volume flow bit by bit by gradually closing valve V2 and take
the measurements according to point 5.
7. Repeat steps 5 and 6 until the volume flow is completely throttled.
8. Additional curves can be recorded with different rotational speed.
GRAPHS:
RESULT:
/Sec.
5. Specific speed corresponding to the flow =……………………

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TABULATION:
S
.
N
o
Rotati
onal
speed
n
(Rev
/ min)
Volu
me
flow
V
(L/mi
n)
Suctio
n
Pressu
re (P1)
(N/m2
)
Discha
rge
Pressu
re (P2 )
(N/m2
)
Temp
eratu
re
(T1)
(0
C)
Electr
ical
Power
(Pel)
(W)
Pressure
Differen
ce (∆P)
(bar)
Head
(H)
(M)
Hydraulic
Power
(Phyd) (W)
Effici
ency
(η)
(%)
Speci
fic
speed
(Ns)
-
1 600 8.7 -0.09 0.98 23.9 147 1.07 17.33 15.515 10.55 0.85
MODEL CALCULATIONS:
1. Phyd= ∆PxV = 1.07x105
x 8.7x0.001/60 =15.515 W
2. H = ∆P/ (ρxg) = 1.07x105
/(1000x9.81) = 17.33m
3. Efficiency = Phyd x100/Pel = 15.515x100/147= 10.55%
= 600 x √(8.7x16.67x10-6
)/ (17.33)3/4
= 0.85
Figure. 7 Gear Oil Pump

23
8. AXIAL FAN
THEORY: An axial fan is a type of a compressor that increases the pressure of the
air flowing through it. The blades of the axial flow fans force air to
move parallel to the shaft about which the blades rotate. In other words, the flow is
axially in and axially out, linearly, hence their name. The design priorities in an
axial fan revolve around the design of the propeller that creates
the pressure difference and hence the suction force that retains the flow across the
fan. The main components that need to be studied in the designing of the propeller
include the number of blades and the design of each blade. Their applications
include propellers in aircraft, helicopters, hovercrafts, ships and hydrofoils. They
are also used in wind tunnels and cooling towers.
AIM:
To identifying characteristics data, to investigate of typical dependencies
and recording the fan characteristics are three aims of our experiment
APPARATUS REQUIRED:
1. Axial fan apparatus setup.
2. Computer system.
FORMULA:
1. ρ =ρ0 x (T0/T1)x(pamb/p0)
ρ0 = 1.293 kg/m3
is air density at reference temperature at T0 = 273.15 K.
p0= 1,013 mbar,in
T1= Temperature of intake air in Kelvin. (1K = X0
C +273.15)
2. Air velocity c = √((2/ρ) x dpN) in m/sec.
dpN = Dynamic pressure in N/m2
.
3. Suction volume flow Vs = c x A in m3
/sec.
A = Area of the intake pipe in m2
.
A= (π/4) x d2
where d = diameter of intake pipe = 110 mm.
4. Power hydraulic Phyd = dpF x Vs in Watts
5. Efficiency = Phydx100/Pel.

24
PROCEDURE:
1. Tare values and enter the ambient pressure in the system diagram.
2. Select ‘Measurement Diagram’ in the program.
3. Enable new series of measurements. Make any setting for the measurement
file.
4. Switch on radial fan, select speed of ____ %
5. For the first measurement, close the throttle valve completely.
6. Wait until the displayed measurement stable. Then record measurements
program is now ready for the next measurement.
7. Open the throttle valve a little bit. The position of the throttle valve is
dependent on the desired number of measurement points. Meaningful
characteristics are often obtained with 5 to 6 measurement points.
8. Repeat the steps until throttle valve is fully open.
9. Repeat the steps with the newly selected speed of 100%, save the
measurements and plot the characteristics curve using system.
OBSERVATION:
Diameter of intake pipe ……110…….. mm
GRAPHS:
Fan speed Vs Efficiency
RESULT:
2. Hydraulic power corresponding to maximum efficiency = …………..W.
3. Electrical power corresponding to maximum efficiency = …………..W
4. Air velocity corresponding to maximum efficiency = …………..m/s
5. Suction volume flow corresponding to maximum efficiency = …….. m3
/Sec

25
TABULATION:
S.
N
o
Fan
speed
(n)
(Rev/
min)
Differe
ntial
Pressu
re flow
(dpN)
(N/m2
)
Pressu
re
increas
e (dpF )
(N/m2
)
Tempe
rature
of
intake
air (T1)
(0
C)
Elect
rical
Powe
r (Pel)
(W)
Ambie
nt
pressur
e (pamb)
(mbar)
Density
of
intake
air (ρ)
(Kg/m3
)
Air
Velo
city
(c)
(m/
sec)
Suction
volume
flow
(Vc)
(m3
/
sec)
Power
hydru
alic
(Phyd)
(W)
Effic
ienc
y (η)
(%)
1 9602 364 394 20.4 78.4 1013 1.203 7.8 0.7412 29.2 37.2
MODEL CALCULATIONS:
1. ρ =ρ0 x (T0/T1)(pamb/p0) = 1.293x (273.15/293.55)x(1013/1013) =1.203kg/m3
.
2. c = √((2/ρ) x dpN) = √((2/1.203) x 36.4) =7.8 m/s
3. Vs = c x A =7.8 x (π/4) x 0.112
=0.07412 m3
/s = 266.8 m3
/hr .
4. Phyd = dpF x Vs = 394x0.07412 =29.2 W
5. Efficiency = Phydx100/Pel = 29.2/78.4 = 37.2%
Main components
M = Drive motor dpF = Differential pressure, radial fan
V-R = Axial fan dpN = Differential pressure, inflow
V1 = Throttle valve n = Speed
Pel = Electrical power of the drive motor T1= Temperature of the intake air

26
9. RADIAL FAN
THEORY: A mine fan (or radial flow fan) in which the air enters along the axis
parallel to the shaft and is turned through a right angle by the blades and
discharged radially. There are three main types with (1) backwardly inclined
blades; (2) radial blades; and (3) forward curved blades. In (2) and (3) the blades
are made of sheet steel, while in (1) the present tendency is to replace curved
sheet-steel blades by blades of aerofoil cross section. The aerofoil bladed radial-
flow fan has an efficiency of about 90%.
AIM:
To determine the efficiency of the radial fan in constant speed condition and
plot the necessary chart.
APPARATUS REQUIRED:
3. Radial fan apparatus setup.
4. Computer system.
FORMULA:
1. Density of intake air ρ =ρ0 x (T0/T1)x(pamb/p0)
ρ0 = 1,293 kg/m3
is air density at reference temperature at T0 = 273.15 K.
p0= 1,013 mbar.
T1= Temperature of intake air in Kelvin. (1K = X0
C +273.15)
2. Air velocity c = √((2/ρ) x dpN) in m/sec.
dpN = Dynamic pressure in N/m2
.
3. Suction volume flow Vs = c x A in m3
/sec.
A = Area of the intake pipe in m2
.
A= (π/4) x d2
where d = diameter of intake pipe = 90 mm.
4. Power hydraulic Phyd = dpF x Vs in Watts

27
PROCEDURE:
1. Tare values and enter the ambient pressure in the system diagram.
2. Select ‘Measurement Diagram’ in the program.
3. Enable new series of measurements. Make any setting for the measurement
file.
4. Switch on radial fan, select speed of ____ %
5. For the first measurement, close the throttle valve completely.
6. Wait until the displayed measurement stable. Then record measurements
program is now ready for the next measurement.
7. Open the throttle valve a little bit. The position of the throttle valve is
dependent on the desired number of measurement points. Meaningful
characteristics are often obtained with 5 to 6 measurement points.
8. Repeat the steps until throttle valve is fully open.
9. Repeat the steps with the newly selected speed of 100%, save the
measurements and plot the characteristics curve using system.
GRAPHS:
Speed in rpm Vs suction volume.
OBSERVATION: Diameter of intake pipe (d) = ……90…….. mm
RESULT:
2. Hydraulic power corresponding to maximum efficiency = …………..W.
4. Air velocity corresponding to maximum efficiency = …………..m/s
5. Suction volume flow corresponding to maximum efficiency = …….. m3
/Sec

28
TABULATION:
S.
N
o
Fan
speed
(n)
(Rev/
min)
Differe
ntial
Pressu
re flow
(dpN)
(N/m2
)
Pressur
e
increas
e (dpF )
(N/m2
)
Tempe
rature
of
intake
air (T1)
(0
C)
Elect
rical
Powe
r (Pel)
(W)
Ambie
nt
pressur
e (pamb)
(mbar)
Density
of intake
air (ρ)
(Kg/m3
)
Air
Velo
city
(c)
(m/
sec)
Suction
volume
flow
(Vc)
(m3
/
sec)
Power
hydru
alic
(Phyd)
(W)
Effici
ency
(η)
(%)
1 2640 22.2 250 25 55.1 1013 1.184 6.12 0.04 10 18.15
MODEL CALCULATIONS:
1. ρ =ρ0 x (T0/T1)(pamb/p0) = 1.293x (273.15/298.15)x(1013/1013) =1.184kg/m3
.
2. c = √((2/ρ) x dpN) = √((2/1.184) x 22.2) =6.12 m/s
3. Vs = c x A =6.12 x (π/4) x 0.092
=0.04 m3
/s = 140.16 m3
/hr .
4. Phyd = dpF x Vs = 250x0.04 =10 W
5. Efficiency = Phydx100/Pel = 10x100/55.1 = 18.15%.
Figure 9. Radial fan
Main components
M = Drive motor dpF = Differential pressure, radial fan
V-R = Radial fan dpN = Differential pressure, inflow
V1 = Throttle valve n = Speed
Pel = Electrical power of the drive motor T1= Temperature of the intake air

29
10. TUBULAR HEAT EXCHANGER (PARRALLEL FLOW)
THEORY: A heat exchanger is a device used to transfer heat between one or more
fluids. The fluids may be separated by a solid wall to prevent mixing or they may
be in direct contact.[1]
They are widely used in space heating, refrigeration, air
conditioning, power stations, chemical plants, petrochemical plants, petroleum
refineries, natural-gas processing, and sewage treatment. The classic example of a
heat exchanger is found in an internal combustion engine in which a circulating
fluid known as engine coolant flows through radiator coils and air flows past the
coils, which cools the coolant and heats the incoming air. There are several types
heat exchanger shell and tube heat exchanger and plate type heat exchanger etc.
AIM: Parameter determination of the tubular heat exchanger in parallel flow of
water.
APPARATUS REQUIRED:
1. Tubular heat exchanger setup.
2. Water tank.
FORMULA:
1. LMTD = Logarithmic Mean Temperature Difference.
LMTD = [Thi –Tci]- [Tho – Tco] / {ln [(Thi- Tci)/(Tho- Tco)]}
Where Tci = T6 = Entry temperature of cold fluid in Kelvin.
Thi = T1 = Entry temperature of hot fluid in Kelvin.
Tco = T4 = Exit temperature of cold fluid in Kelvin.
Tho = T3 = Exit temperature of hot fluid in Kelvin.
2. Qh = Heat transfer rate from hot water in KJ = mh x Cph [Thi – Tho]
Where mh = Mass flow rate of hot water [Kg/s]
Cph = Specific heat of hot water [KJ/KgK] = 4.187 KJ/KgK
3. Qc = Heat Transfer rate to the cold water = mc x Cpc [Tco- Tci]
Where mc = Mass flow rate of cold water [Kg/s]
Cpc = Specific heat of cold water [KJ/KgK] =4.187 KJ/KgK
4. Q = Heat transfer rate in Watts = [Qh + Qc] / 2
5. U = Overall Heat transfer co-efficient W/m2K = Q/(A x[ΔT]M)

30
Where [ΔT]M = LMTD
A = Area = πdl
6. Cr = Cmin/ Cmax
Ch = Cph x mh
Cc = Cpc x mc
In Cc , Ch which is minimum called Cmin and which is maximum called Cmax.
7. NTU = No of transfer units = Ux A/ Cmin
1- exp [ - NTU x (1+ Cr)]
8. Effectiveness E = ------------------------------------
1+ Cr
PROCEDURE:
1. Give the necessary connection to the set up.
2. Heat the water in the setup using heater.
3. Give the flow of hot water and cold water using the valve according to the
diagram. Note down the flow rate of hot and cold water.
4. Now the change in temperatures take place, note down the temperatures
after the change in temperatures reaches a steady value
5. Repeat the process of for other flow rates.
6. Tabulate the value and plot the graph.
GRAPHS: Heat transfer rate Vs Effectiveness.
Observation:
Overall length (L) = ……..560…….. mm.
Diameter (D) = ………..7…….. mm.
RESULT:
1. Heat transfer rate (Q)= _______________ W
2. Overall heat transfer coefficient (U)= _____________ W/m2
k.
3. Effectiveness (E) = ___________ .

31
TABULATION:
S
.
N
o
Flow rate
of Hot
water
Flow rate
of Cold
water
Inlet
temp of
hot
water
(Thi)
(T1)
outlet
temp of
hot water
(Tho) (T3)
Inlet
temp of
cold
water
(Tci)
(T6)
Outlet
temp of
hot water
(Tco)
(T4) LMT
D
Hea
t
tran
sfer
rate
(Q)
Ove
r all
heat
tran
sfer
coef
ficie
nt
(U)
Effe
ctiv
enes
s
(E)
L/hr Kg /s
L/mi
n
Kg /s 0
C K 0
C K 0
C K 0
C K
1 144 .04 120 .0333 56 329 45 318 34 307 39 312 12.31 1.27 8.4 .403
MODEL CALCULATIONS:
LMTD = [Thi – Tci] - [Tho – Tco] / ln [Thi – Tci/Tho – Tco]
= [329 – 307] – [318 – 312] / ln [(329 – 307) / (318 – 312)]= 12.31 K.
Qh = mh x Cph [Thi – Tho] = 0.04 x 4.187 x [329 – 318] = 1.842 KJ/sec.
Qc = mc x cpc [Tco –Tci]= 0.0333 x 4.187 [312 – 307] = 0.691 KJ/sec.
Q = [Qh + Qc] / 2 = [1.842 + 0.691] / 2 = 1.27 KJ/sec.
A = π x D x L= π x 0.007 x 0.56= 0.0123 m2
.
U = Q/(A x[ΔT]M)= 1.27 / (0.0123 x 12.31)= 8.4 W/m2
K.
Cr = Cmin/ Cmax = 0.14/0.167= 0.8383
Ch = Cph x mh = 4.187 x 0.04= 0.167 = Cmax
Cc = Cpc x mc = 4.187 x 0,0333= 0.140 = Cmin

32
NTU = Ux A/ Cmin = 8.4x0.0123/ 0.14 = 0.737.
1- exp [ - NTU x (1+ Cr)] 1- exp [- 0.737x 1.8383]
9. E = ---------------------------------- = --------------------------------- = 0.4036
1+ Cr 1.8383
Fig 12.Parrallel Flow Tubular Heat Exchanger

33
11. TUBULAR HEAT EXCHANGER (COUNTER FLOW)
THEORY: A heat exchangers are classified parallel flow and counter flow based
on the direction of both the fluids flow. In parallel flow heat exchanger both the
fluids (hot and cold) are flowing in the same direction. But in counter flow heat
exchanger both the fluids are flowing in the opposite direction.
AIM: Parameter determination of the tubular heat exchanger in counter flow of
water.
APPARATUS REQUIRED:
1. Tubular heat exchanger setup.
2. Water tank.
FORMULA:
1. LMTD = Logarithmic Mean Temperature Difference.
LMTD = [Thi –Tci]- [Tho – Tco] / {ln [(Thi- Tci)/(Tho- Tco)]}
Where Tci = T4 = Entry temperature of cold fluid in Kelvin.
Thi = T1 = Entry temperature of hot fluid in Kelvin.
Tco = T6 = Exit temperature of cold fluid in Kelvin.
Tho = T3 = Exit temperature of hot fluid in Kelvin.
2. Qh = Heat transfer rate from hot water in KJ = mh x Cph [Thi – Tho]
Where mh = Mass flow rate of hot water [Kg/s]
Cph = Specific heat of hot water [KJ/KgK] = 4.187 KJ/KgK
3. Qc = Heat Transfer rate to the cold water = mc x Cpc [Tco- Tci]
Where mc = Mass flow rate of cold water [Kg/s]
Cpc = Specific heat of cold water [KJ/KgK] =4.187 KJ/KgK
4. Q = Heat transfer rate in Watts = [Qh + Qc] / 2
5. U = Overall Heat transfer co-efficient W/m2K = Q/(A x[ΔT]M)
Where [ΔT]M = LMTD
A = Area = πdl
6. Cr = Cmin/ Cmax
Ch = Cph x mh

34
Cc = Cpc x mc
In Cc , Ch which is minimum called Cmin and which is maximum called Cmax.
7. NTU = No of transfer units = Ux A/ Cmin
1- exp [ - NTU x (1- Cr)]
8. Effectiveness E = -------------------------------------------
1- {Cr xexp [ - NTU x (1- Cr)]}
PROCEDURE:
1. Give the necessary connection to the set up.
2. Heat the water in the setup using heater.
3. Give the flow of hot water and cold water using the valve according to the
diagram. Note down the flow rate of hot and cold water.
4. Now the change in temperatures take place, note down the temperatures
after the change in temperatures reaches a steady value
5. Repeat the process of for other flow rates.
GRAPHS: Heat transfer rate Vs Effectiveness.
Observation:
Overall length (L) = ……..560…….. mm.
Diameter (D) = ………..7…….. mm.
RESULT:
1. Heat transfer rate (Q)= _______________ W
2. Overall heat transfer coefficient (U)= _____________ W/m2
k.
3. Effectiveness (E) = ___________ .

35
TABULATION:
S.
N
o
Flow rate
of Hot
water
Flow rate
of Cold
water
Inlet
temp of
hot
water
(Thi)
(T1)
outlet
temp of
hot
water
(Tho)
(T3)
Inlet
temp of
cold
water
(Tci)
(T4)
Outlet
temp of
hot
water
(Tco)
(T6)
LMT
D
Hea
t
tran
sfer
rate
(Q)
Over
all
heat
trans
fer
coeff
icien
t (U)
Effe
ctiv
enes
s
L/h
r
Kg /s L/hr Kg /s 0
C K 0
C K 0
C K 0
C K
1 288 .08 191 .053 82 355 57 330 35 308 51 324 26.24 5.95 18.43 0.44
MODEL CALCULATIONS:
LMTD = [Thi – Tci] - [Tho – Tco] / ln [Thi – Tci/Tho – Tco]
= [355 – 324] – [330 – 308] / ln [(355 – 324) / (330 – 308)]= 26.24 K.
Qh = mh x Cph [Thi – Tho] = 800 x 10-4
x 4.187 x [355 – 330] = 8.347 KJ/sec.
Qc = mc x cpc [Tco –Tci]= 0.053 x 10-4
x 4.187 [32 4 – 308] = 3.551 KJ/sec.
Q = [Qh + Qc] / 2 = [8.347 + 3.551] / 2 = 5.95 KJ/sec.
A = π x D x L= π x 0.007 x 0.56= 0.0123 m2
.
U = Q/(A x[ΔT]M)= 5.95 / (0.0123 x 26.24)= 18.4352 W/m2
K.
Cr = Cmin/ Cmax = 0.335/0.222 = 1.51
Ch = Cph x mh = 4.187 x 0.08 = 0.335 = Cmax
Cc = Cpc x mc = 4.187 x 0.053 = 0.222 = Cmin
NTU = Ux A/ Cmin = 18.4352x0.0123/ 0.222 = 1.02.

36
1- exp [ - NTU x (1- Cr)] 1- exp [- 1.02 x -0.51]
E = -------------------------------------- = -------------------------------------- = 0.44
1- {Cr xexp [ - NTU x (1- Cr)]} 1- {1.51xexp [- 1.02 x -0.51]}
Fig 13. Counter Flow Tubular Heat Exchanger

37
12. AIR COMPRESSOR
THEORY: An air compressor is a device that converts power (using an electric
motor, diesel or gasoline engine, etc.) into potential energy stored in pressurized air
(i.e., compressed air). This air compressor is a two stage reciprocating type. The air
is sucked from atmosphere and compressed in the first cylinder. The compressed
air then passes through an inter cooler into the second stage cylinder, where it is
further compressed. The compressed air then goes to a reservoir through a safety
valve. This valve operates an electrical switch that shuts off the motor when the
pressure exceeds the set limit.
AIM: To conduct a performance test on a two stage air compressor and determine
its volumetric efficiency.
APPARATUS REQUIRED:
1.Two stage air compressor apparatus.
2. Stop watch
FORMULA:
1. Volume V0 = Ad x√(2x∆p/ρ) ) in m3
.
Ad = Area of the duct= 1.131x10-4
m2
.
ρ = Density of air =1.293 kg/m3
.
(1bar = 1x105
N/m2
) (1mbar = 1x10-3
bar) (1Pa = 1N/m2
)
2. Isothermal power Piso = p1V0 ln (p4/p1)
3. Efficiency = Piso/Pel.
1 p1 - Inlet pressure
2 T1 - Inlet temperature
3 p2 - Pressure after 1st compressor stage
4 T2-Temperature after 1st compressor stage
5 p4-Pressure vessel pressure
6 T3-Temperature before 2nd compressor stage
7 ∆p-Differential pressure across Venturi nozzle
8 T4-Temperature after 2nd compressor stage

38
PROCEDURE:
1. Close the outlet valve.
2. Switch on compressor (see Section 2.4) If it does not start up, it is possible
that the over-current protection switch may have cut out directly on the
motor - restart.
3. Allow the system to run, until a constant pressure p3 has built up, set the
desired final pressure with the bleeder valve and record the measured values.
GRAPHS:
Pressure Vs Volume
OBSERVATION: Area of duct inlet= ……1.131x10-4
m2
…….
RESULT:
1. Maximum efficiency of the compressor = ……………………… %
2. Isothermal power corresponding to maximum efficiency = …………..W.
4. Volume flow rate V0
at maximum efficiency = …………………l/min

39
TABULATION:
S.
No
p1 T1 p2 T2 T3 p4 T4 ∆p V0
Pelec Tim
e
Piso η
Un
it
bar 0
C bar 0
C 0
C bar 0
C mba
r
m3
/ Sec W min W %
1 0.99 23 3.4 127.1 58.8 11.7 151.6 7.4 38.26x10-4
2550 3 935.43 36.7
MODEL CALCULATIONS:
1. V0 = Ad x√(2x∆p/ρ) ) = 1.131X10-4
x(√(2x7.4/1.293) ) = 38.26x10-4
m3
/s.
2. Piso = p1V0 ln (p4/p1) = 0.99x105
x38.26x10-4
xln(11.7x105
/0.99x105
) =935.43W
3. η = Pisox100/Pel = 935.43x100/2550 =36.7%.
Fig 12 Air compressor

40
13. MULTI PURPOSE AIRDUCT
THEORY: There are various forms of heat transport: Convection is the transport
of heat by a moving fluid. Example: In forced convection, a conveying unit (pump,
blower) moves the fluid to be heated or cooled along the surfaces of a heat
exchanger. Thermal radiation is energy emitted by electromagnetic waves.
Example: Thermal radiation from the sun. Conduction is kinetic energy being
transported between two neighboring atoms or molecules. Example: A refrigerator
is insulated to prevent conduction.
Convective heat transfer takes part in heat exchangers and plays a large role in
many areas of industry. There are many different forms of heat exchangers which
transfer heat from one medium to another. Convective heat transfer in heat
exchangers can take place according to different principles: Parallel flow,
Counterflow, Cross-flow.
The Multipurpose Air Duct and Heat Transfer Unit WL 312 offers an excellent
supplement for calculations. It can be used to determine convective heat transfer on
an experimental basis. It provides a view of industrial applications with the
possibility of installing different types of heat exchangers with different heat
transfer media.
AIM: To determine the flow velocity and volume flow rate in multipurpose air
duct.
APPARATUS REQUIRED:
1. Air duct compressor apparatus.
2. Stop watch
FORMULA:
1. Velocity c = √(2xpdyn/ρ) ) in m/sec.
pdyn = Differential pressure between the ambient air and air duct (PD1) in mbar.
ρ = Density of air =1.293 kg/m3
.
(1bar = 1x105
N/m2
) (1mbar = 1x10-3
bar) (1Pa = 1N/m2
)
2. Volume flow rate V = c xAd in m3
/sec.
Ad= Area of duct inlet in m2
= hdx bd.
hd = Height of the duct in mm.
bd = Width of the duct in mm.

41
PROCEDURE:
1. Place the throttle valve in a position that is vertical (90°) to the air flow. This
ensures that the maximum possible flow rate of the fan is achieved, since the
resistance is at its lowest level on the pressure side.
2. Switch on the fan.
3. Read the dynamic pressure (which is a proportion of the flow velocity in the air
duct) on the digital display with the differential pressure sign (flow).
4. Change the position of the throttle valve in order to obtain a different dynamic
pressure (flow rate).
GRAPHS:
Pressure Vs Velocity
OBSERVATION: hd = Height of the duct …290…. mm.
bd = Width of the duct …150…mm.
RESULT:
1. Maximum Velocity of air in duct = ……………. m/s
2. Maximum flow rate of air in duct=……………... m3
/s
3. Angle of opening of throttle valve at Maximum flow rate of air in
duct=……………... 0

42
TABULATION:
S.No Throttle valve
Position in °
Dynamic
pressure
pdyn
in mbar
Flow velocity
c
in m/s
Flow rate
V in m³/Sec
1 90 0.873 12.1 0.505
MODEL CALCULATIONS:
4. Ad = hdx bd = 0.29x0.15= 0.0435 m2
.
5. c = √(2xpdyn/ρ) ) = (√(2x0.873x105
x10-3
/1.293) ) = 11.62 m/s.
6. V = c xAd = 11.62x0.0435=0.505 m3
/sec.
Switch Cabin Test section
Throttle valve Fig 13. Air duct Air inlet

43
IMPORTANT TERMS
1. 1liter =1000 cm3
= 0.001 m3
2. 1bar = 1x105
N/m2
3. 1liter/min =16.67x10-6
m3
/Sec
4. 1liter = 1kg
5. 1Pa = 1N/m2
.
6. 1mbar =0.001bar

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